How do you differentiate $f (x) = \sqrt{{(x^{3} + x^{2})}^{- 1} + x^{3}}$ $f(x)=sqrt((x^3+x^2)^-1+x^3)$ using the chain rule?

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Answer 1 · 2018-07-13T12:13:06

$f'(x)=(-(x^3+x^2)^(-2)(3x^2+2x)+3x^2)/sqrt((x^3+x^2)^(-1)+x^3)$

We Need the chain rule

$(f(g(x))'=f'(g(x))*g'(x)$

so we get

$f'(x)=1/2*((x^3+x^2)^(-1)+x^3)^(-1/2)*(-(x^3+x^2)^(-2)(3x^2+2x)+3x^2)$

this is
$((-(x^3+x^2)^(-2)(3x^2+2x)+3x^2))/sqrt((x^3+x^2)^(-1)+x^3)$

How do you differentiate f(x)=sqrt((x^3+x^2)^-1+x^3) f(x)=√(x3+x2)−1+x3f(x)=sqrt((x^3+x^2)^-1+x^3) using the chain rule?